I was trying the following problem :
Let $\operatorname{O}(n)$ be the group of $n \times n$ orthogonal matrices. Show that if $\gamma : (-1,1) \to \operatorname{O}(n)$ is a differentiable curve such that $\gamma(0) = I_n$ , then $\gamma'(0)$ is a skew-symmetric matrix.
What can be said if $\gamma(0)$ is an arbitrary element $O \in \operatorname{O}(n)$?
Repeat the exercise replacing $\operatorname{O}(n)$ by $\operatorname{SO}(n)$.
I have no ideas regarding the problem and hence could not proceed.Thanks in advance for help.
If $M,N$ are smooth manifolds and $f \colon M \to N$ is a smooth map, then its pushforward at $p \in M$ is $$f_{p*}\colon T_pM \rightarrow T_{f(p)}N :X_p \mapsto \frac{d}{dt}f(\alpha(t))\biggr\lvert_{t=0}$$ where $\alpha$ is a curve on $M$ such that $\alpha(0) = p$ and $\alpha'(0) = X_p$.
In the case where $f$ is the curve $\gamma$, then $M=(-1,1)$, $N=\mathrm{O}(n)$, $T_pM = \mathbb{R}$ and $T_{f(p)}N \subseteq M(n,\mathbb{R})$ (the space of all $n \times n$ matrices). So $\gamma(t) \in \mathrm{O}(n)$ for any $t \in (-1,1)$, hence $\gamma(t)^T \gamma(t) = \mathrm{id}_n$ (where $^T$ denotes the transpose). Check that its pushforward at the point $t$ is the derivative $\gamma'(t)$. Since the identity map is constant, you can compute $$\frac{d}{dt} \gamma(t)^T \gamma(t)\biggr\lvert_{t=0} = 0 = \gamma'(0)^T\gamma(0) +\gamma(0)^T\gamma'(0).$$ Now $\gamma(0) = \mathrm{id}_n = \gamma(0)^T$, then you get $\gamma'(0)^T+\gamma'(0) = 0$, that is $\gamma'(0)^T = -\gamma'(0)$. Thus $\gamma'(0)$ is a skew-symmetric matrix in $M(n,\mathbb{R})$.
Now you should be able to complete the remaining points.